Question

A body of mass $2\text{kg}$ initially at rest moves under the action of an applied horizontal force of $7\text{N}$ on a table with coefficient of kinetic friction ${\mu }_k=0.1$.
Calculate the work done by kinetic friction in $10\text{sec}(g=9.8{\text{m/s}}^2)$

• A. $246.96\text{J}$
• B. $-246.96\text{J}$
• C. $882\text{J}$
• D. $-882\text{J}$

Answer 1

The correct option is B $-246.96\text{J}$

By the $FBD$ of the block,


$N=mg=2\times 9.8=19.6\text{N}$

$\therefore$ Kinetic friction $={\mu }_{k}N$
$=0.1\times 19.6$
$=1.96\text{N}$

Now.
$F_{net}=F-f_k=ma$
$\Rightarrow 7-1.96=2\times a$
Retardation due to friction $\Rightarrow a=2.52{\text{m/s}}^2$
$\therefore$ Distance moved, $s=\frac{1}{2}a{t}^2=\frac{1}{2}\times 2.52\times {10}^2=126\text{m}$

$\therefore$ work done by kinetic friction $=-F.s=-1.96\times 126$
$=-246.96\text{J}$,
Here the work done is negative as the force and displacement are in opposite direction

Hence option B is the correct answer